Given that \(tan^{-1}(x) + cot^{-1}(x) = \frac{2π}{2}\), the equation can be rearranged as:
\((tan^{-1}(x) + cot^{-1}(x))^2 - 2tan^{-1}(x)(\frac{2π}{2} - tan^{-1}(x)) = \frac{8}{(\frac{5π}{2})}\)
Simplifying further: \(2(tan^{-1}(x))^2 - 2(\frac{2π}{2})tan^{-1}(x) - \frac{8}{(\frac{3π}{2})} = 0\)
This can be rewritten as: \(2(tan^{-1}(x))^2 - 2(π)tan^{-1}(x) - \frac{8}{(\frac{3π}{2})} = 0\)
From this equation, we can deduce that \(tan^{-1}(x) = \frac{-4π}{3}\) or \(\frac{4π}{3}\).
Hence, the solutions are \(x = -1\).
The range of the real valued function \( f(x) =\) \(\sin^{-1} \left( \frac{1 + x^2}{2x} \right)\) \(+ \cos^{-1} \left( \frac{2x}{1 + x^2} \right)\) is:
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)